Question

What fraction of the α particles in Rutherford's gold foil experiment are scattered at large angles? Assume the gold foil is two layers thick, as shown in the figure, and that the approximate diameters of a gold atom and its nucleus are 1.5 Å and 1.5×10−4Å,

Answer 1

Ans. The atom is assumed to be 2D structure because a-particles are bombarded from one direction only (say, perpendicular to the plane of sheet).

Radius of gold atom = diameter / 2 = 1.5A / 2 = 0.75 A

                        Area of a gold atom = (pi) r² = 3.14158 x (0.75 A) = 1.767 A²

Radius of nucleus = (1.5 x 10^-4 A) / 2 = 0.75 x 10^-4 A.

                        Area of nucleus = 3.14159 x (0.75 x 10^-4 A) = 1.767 x 10^-8 A²

The positively charged a-particles deflect at larger angles when it collides with the positively charged nucleus.

In multilayered structures, the nuclei are arranged side-by-side (NOT exactly on top or below) to minimize repulsion. So, increasing the number of layers also increases the active area of the nucleus. It also means that, increasing the number of atom-layers in gold sheet also increases the fraction of a-particles scattered as large angles.

So,

Functional area of nuclei in 2-layered sheet = 2 x area of nucleus on one-layer

                                                                        = 2 (1.767 x 10^-8 A²)

Now,

Fraction of a-particles scattered at large angle = Fraction of nuclei area in gold sheet

                                                            = Functional area of nuclei / Area of Atom

                                                            = [2 x 1.767 x 10^-8 A²] / 1.767 A²

                                                            = 2.0 x 10^-8

                                                            = 2 / (10⁸)

That is, nucleus constitutes around 2.0 parts in every 10⁸ part of the gold atom.

So, 2 a-particles in every total 10⁸ particles would scatter at large angles. Or, fraction of a-particles scattered at large angles = 2 / 10⁸